document.write( "Question 871846: A picture has dimensions 20cm by 30 cm. It is surrounded by a frame of uniform width whose outer edge is a rectangle with an area of 1800 cm squared. To the nearest .1 cm, find the width of the frame. \n" ); document.write( "

Algebra.Com's Answer #525772 by mananth(16946) $\"\"$ $\"About$

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Let the width be x
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\n" ); document.write( "Length of picture 30 cm 54
\n" ); document.write( "width of picture 20 cm 44
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\n" ); document.write( "Area = 600 m^2
\n" ); document.write( "Area of frame 1800 m^2
\n" ); document.write( "length of frame & plot 30 + 2 x
\n" ); document.write( "width of frame & plot 20 + 2 x
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\n" ); document.write( "( 30 + 2 x ) ( 20 + 2 x ) + 0 = 1,800
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\n" ); document.write( " 600 + 60 x + 40 x + 4 X^2 + 0 = 1,800
\n" ); document.write( " 4 X^2 + 100 x + -1,800 = 0
\n" ); document.write( " Find the roots of the equation by quadratic formula
\n" ); document.write( " a= 4 b= 100 c= -1,800
\n" ); document.write( " b^2-4ac= 10,000 - -28,800
\n" ); document.write( " b^2-4ac= 38,800 $\"sqrt%28%0938%2C800%09%29\"$ = 197
\n" ); document.write( " $\"x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29\"$
\n" ); document.write( " $\"x1=%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29\"$
\n" ); document.write( " x1=( -100 + 197 )/ 8
\n" ); document.write( " x1= 12.122
\n" ); document.write( " x2=( -100 -197 ) / 8
\n" ); document.write( " x2= -15.500
\n" ); document.write( " Ignore negative value
\n" ); document.write( " width = 12.1 cm
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